If every pair of non empty open sets have non empty intersection, then it is hyperconnected Prove that, in a topological space if every pair of non empty open sets have non empty intersection, then it is hyperconnected.
Steen and Seebach has already proved this theorem. But I failed to search and find their paper online. The converse part of this theorem is known to me.

A topological space $X$ is said to be hyperconnected if closure of each open set is $X$.

 A: The following are equivalent for any space $X$:


*

*Any two open non-empty subsets of $X$ have non-empty intersection.

*$X$ cannot be written as the union of two proper (i.e $\neq X$) closed subsets.

*Every non-empty open set of $X$ is dense (i.e $\overline{U} = X$).

*Every proper closed set has empty interior.


Proof:


*

*implies 2. : Assume 1. Suppose $X = C\cup D$,where $C,D$ are closed.
Then $(X\setminus C) \cap (X\setminus D) = X\setminus (C \cup D) = \emptyset$ by de Morgan, so 1. implies that $X\setminus C$ or $X\setminus D$ is empty (they are open, so cannot both be non-empty). So $C=X$ or $D=X$.

*implies 3.: Suppose $O$ is non-empty open. If $C = \overline{O} \neq X$, then $X = C \cup (X \setminus O)$ writes $X$ as a union of two proper closed sets, which cannot be.

*implies 4.: If $C$ is closed, $C \neq X$, and suppose that $O:= \operatorname{int}(C) \neq \emptyset$. Then by 3. $X =\overline{O} \subseteq \overline C = C \neq X$, contradiction. So 4. holds.

*implies 1.: Suppose $U$ and $V$ are non-empty open subsets of $X$. If $U \cap V = \emptyset$, $U \subseteq X\setminus V$, so $\emptyset \neq U \subseteq \operatorname{int}(X \setminus V)$, contradiction with 4. 
A: Let $U \subset X$ be open and non-empty. If $\overline U \neq X$, then $X - \overline U$ is a non-empty open set that doesn't intersect $U$; this contradicts your hypothesis.
